Solid state | Practice questions | CBSE 12

1. If  Radius (r) = 125 pm and  z = 4 (FCC),find the Edge length of unit cell (a) and Number of unit cells in 1 cm3 of Al = ?  

      Answer: 2.266 × 1024 unit cells

2. If Edge length (a) = 3.9231 Å or 3.9231 × 10−8 cm,  Atomic weight of Pt (M)= 195.08  and 

      z = 4 (FCC) find the Density (d) and Atomic radius (r) = ? 

      Avogadro’s number = NA = 6.023 × 1023 

      Answer:  r = 138 .7 pm

3. Edge length (a) = 4.049 Å, Atomic mass of Al (M) = 26.98 g/mol and z = 4 (FCC) find Density (d) = ? 

     Avogadro’s number = NA = 6.023 × 1023

     Answer: 2.699 g/cc

4. A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and that of P are at the bodycentre. What is the formula of the compound? What are the coordination numbers of P and Q? 

     Answer: The formula of the compound is PQ.The coordination number of both P and Q is 8.

5. The edge length of the unit cell of Ta, is 330.6 pm; the unit cell is body-centered cubic. Tantalum has a density of 16.69 g/cm3 (a) calculate the mass of a tantalum atom. (b) Calculate the atomic weight of tantalum in g/mol.

      Answer: 3.015 x 10¯22 g and 181.6 g/mol 

6. An element with density 11.2 g cm-3 forms an FCC lattice with edge length of 4 X 10-8cm. Calculate the atomic mass of the element. (Given: N = 6.022 X 1023 mol-1)

        Answer: 107.9 g 

7. Tungsten crystallizes in a body-centred cubic unit cell. If the edge of the unit cell is 316.5 pm, what is the radius of the tungsten atom?

       Answer: 137.04 pm

8. The well known mineral fluorite is chemically calcium fluoride. It is known that in one unit cell of this mineral there are 4 Ca2+ ions and 8 F–. ions and that Ca2+ ions are arranged in a FCC lattice. The F- ions fill all tetrahedral holes in the face centred cubic lattice of Ca2+ ions. The edge of the unit cell is 5.46 X 10-8 cm in length. The density of the solid is 3.18 g cm-3. Use this information to calculate Avogadro’s number. (Molar mass of CaF2 = 78.08 g mol-1)

       Answer: 6.023 X 10 23 

9. What is the formula of a compound in which the element ‘Y’ form hep lattice and atoms of ‘X’ occupy 1/3 r“ of octahedral voids.

             Answer: XY3 

10. What is the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy 2/3rd of tetrahedral voids?

Answer: X4 Y1

11. Calculate the packing efficiency of a metal crystal for simple cubic lattice. 

              Answer: 52.4%

12. Why does Li Cl acquire pink colour when heated in Li vapours?

13. What type of semiconductor is obtained when silicon is doped with boron?

14. Express the relationship between atomic radius(r) and the edge length (a) of the FCC unit cell.

15. What is the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy l/3rd of tetrahedral voids? 

            Answer : X ⅔ : Y 

Solutions | Raoult's law | Solved problems

QUESTION 1:

Calculate the osmotic pressure of a solution 0.1 molar of sucrose (Mw = 342) at  20oC.

SOLUTION:  

From the formulae,  

Molality = moles/kilogram of solvent

Molarity = moles/liter of solution

The grams of sucrose in the solutions are 34.2 and the total weight of the solution is

34.2 + 1000 grams of solvent = 1034.2

Assuming a density is 1 

          0.1 moles : 1034.2 = M : 1000 grams of solution

          M = 100/1034.2 = 0.097

          p = (0.097)(0.0820578 liter atm mol-1 K-1)(293) = 2.34 atm

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QUESTION 2:

A mixture of water and acetone at 756 mm boils at 70oC. Calculate the percentage composition of the  mixture using the following table:

Temperature oC	Vapor pressure (atm) Acetone	Vapor pressure (atm) Water
60	1.14	0.198
70	1.58	0.312
80	2.12	0.456
90	2.81	0.694

SOLUTION:  

According to Raoult's law, 

P = xacetone Po acetone + xwater Powater 

    = 756/760 =0.995 atm

Substituting the values at 70oC ,  

P = xacetone 1.58 + xacqua 0.312 = 0.995

We also know that, 

xacetone + xacqua = 1

So, xacqua = 1 - xacetone

  So,  xacetone 1.58 + 0.312 (1 - xacetone) = 0.995

xacqua 1.58 + 0.312 - 0.312 xacetone = 0.995

xacetone 1.26+ 0.312 = 0.995

xacetone = (0.995 - 0.312) / 1.26 = 0.54

So,  xwater = 1 - 0.54 = 0.46

Considering that we have 100 moles of solution, 54 moles of  acetone and 46 moles of water is present in the sample. 

54 moles of  acetone is 54 x 58 =3132 grams

and 46 moles of water is 46 x 18 = 828 grams

So,  % water = (828)(100)/3132 + 828) = 21%

       % acetone = 100-21 = 79%

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QUESTION 3:

The addition of 114 grams of sucrose to 1000 grams of water lowers the vapor pressure of water from 17.540 to 17.435. Calculate the molecular weight of sucrose.

SOLUTION:  

The difference in pressure: 17.540 - 17.435 = 0.105 

Then, 

0.105 = 17.540 x2  = nsolute/(nsolvent + nsolute)

nsolvent = 1000/mw water = 1000/18 = 55.55 moles

0.105 = 17.540 x2  = 17.540 nsolute/(55.55 + nsolute)

by solving this,

         nsolute = 0.335 moles and 

MW = 114/0.335 = 340

The calculated molecular weight (340) is very close to the true molecular weight of sucrose 342

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QUESTION 4:

Calculate the osmotic pressure at 20o of a suspension containing 60g/l of solid particles each particle having a mass of 10-9 grams (1 nanogram).

SOLUTION:   

Formula:  p = MRT ( T = 278 + 20 = 298 oK and R = 0.0823)

The number of particles in one liter of solution is = 60/10-9 = 6 x 1010

1 mole of any substance contains an Avogadro number of particles (6 x 1023)

The number of moles in one liter of suspension = 6 x 1010/6 x 1023 =  10-13

which is equal to the molarity of the suspension.

So, p = (0.0823)(298) (10-13) = 24.5 x 10-13 atm

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QUESTION 5:

Calculate the vapor pressure of a mixture containing  252 g of n-pentane (Mw = 72) and 1400 g of n-heptane (Mw = 100) at 20oC. The vapor pressure of n-pentane and n-heptane are 420 mm Hg and 36 mm Hg respectively.

SOLUTION: 

From  Raoult's law, P = Po x

Here,

P = vapor pressure of the component in the mixture.

Po= vapor pressure of the pure component.

x = Molar fraction of the component in the mixture.

Calculation of mole fraction: 

Mole fraction = moles of one component / Total moles in solution

Moles n-pentane = 252/72 = 3.5

Moles n-eptano = 1400/100 = 14

Totals = 3.5 + 14 = 17.5 moles

xn-pentane = 3.5/17.5 = 0.2 ,  xn-eptane = 14/17.5 = 0.8

So,  Pn-pentane =  0.2 x 420 = 84 mm Hg ; Pn-eptane =  0.8 x 36 = 28.8 mm Hg

and the vapor pressure of mixture is, Pmixture = 84 + 28.8 = 112.8 mm

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QUESTION 6:

Calculate the boiling point (at 1 atm) of a solution containing 116 g of acetone (Mw = 58) and 72 g of water (Mw = 18) by using the following table:

Temperature oC	Vapor pressure (atm) Acetone	Vapor pressure (atm) Water
60	1.14	0.198
70	1.58	0.312
80	2.12	0.456
90	2.81	0.694

SOLUTION: 

A liquid starts to boil when its vapor pressure matches the atmospheric pressure (1 atm in this case).

According to Raoult's law

P = xacetone Po acetone + xwater Powater = 1 atm

Calculate the molar fractions: 

Mole fraction = moles of one component / Total moles in solution

molesacetone = 116/58 = 2 ; moleswater = 72/18 = 4

So, Total moles = 2 + 4 =  6 

Substituting the given values, 

xacetone = 2/4 = 1/3  ;  xwater = 4/6 = 2/3

So,  P = 2/3 Po acetone + 1/3 Powater= 1 atm

By trials, using the table, we can find the values of vapor pressure which satisfies the above equation. The best result is obtained by using  the values at  80oC :

P = 2/3 0.456 + 1/3 2.12 = 1.01 atm

So, the boiling point is about 80oC.